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If i(t) = 4 sin 2t + 3t, find i(t) when t = pi/3 and when t = 0.6 rad.

Mathematics

ScanSolve AI Answer

Answer

6.6057
5.5282
Graph of i(t) = 4sin(2t) + 3t

Use the left and right arrow keys to trace the curve, up and down to jump between key points, and plus and minus to zoom.

Key points of i(t) = 4sin(2t) + 3t
Pointty
Maximum−5.31−12.21
Minimum−4.12−16.07
Maximum−2.16−2.78
Minimum−0.98−6.64
x-intercept00
Maximum0.986.64
Minimum2.162.78
Maximum4.1216.07
Minimum5.3112.21

Step-by-step solution

2 steps

Here's how to find the values of i(t)i(t) for the given times.

Given the function i(t)=4sin⁡2t+3ti(t) = 4 \sin 2t + 3t.

  1. Find i(t)i(t) when t=π3t = \frac{\pi}{3}.

    Substitute t=π3t = \frac{\pi}{3} into the function:

    i(π3)=4sin⁡(2⋅π3)+3⋅π3i\left(\frac{\pi}{3}\right) = 4 \sin\left(2 \cdot \frac{\pi}{3}\right) + 3 \cdot \frac{\pi}{3} i(π3)=4sin⁡(2π3)+πi\left(\frac{\pi}{3}\right) = 4 \sin\left(\frac{2\pi}{3}\right) + \pi

    We know that sin⁡(2π3)=sin⁡(π−π3)=sin⁡(π3)=32\sin\left(\frac{2\pi}{3}\right) = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}.

    i(π3)=4(32)+πi\left(\frac{\pi}{3}\right) = 4 \left(\frac{\sqrt{3}}{2}\right) + \pi i(π3)=23+πi\left(\frac{\pi}{3}\right) = 2\sqrt{3} + \pi

    Now, calculate the numerical value:

    i(π3)≈2(1.73205)+3.14159i\left(\frac{\pi}{3}\right) \approx 2(1.73205) + 3.14159 i(π3)≈3.46410+3.14159i\left(\frac{\pi}{3}\right) \approx 3.46410 + 3.14159 i(π3)≈6.6057i\left(\frac{\pi}{3}\right) \approx \boxed{6.6057}
  2. Find i(t)i(t) when t=0.6t = 0.6 rad.

    Substitute t=0.6t = 0.6 into the function:

    i(0.6)=4sin⁡(2⋅0.6)+3⋅0.6i(0.6) = 4 \sin(2 \cdot 0.6) + 3 \cdot 0.6 i(0.6)=4sin⁡(1.2)+1.8i(0.6) = 4 \sin(1.2) + 1.8

    Using a calculator (ensure it's in radian mode):

    sin⁡(1.2)≈0.932039\sin(1.2) \approx 0.932039 i(0.6)≈4(0.932039)+1.8i(0.6) \approx 4(0.932039) + 1.8 i(0.6)≈3.728156+1.8i(0.6) \approx 3.728156 + 1.8 i(0.6)≈5.5282i(0.6) \approx \boxed{5.5282}
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